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Proof step 3 from proof union-subset-inj (previous step)

⊦ (∀ a. count(a, bag ∪ bag1) ≤ count(a, bag ∪ bag2)) ↔ bag1 ⊆ bag2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: union: ⊦ count(a, bag0 ∪ bag1) = count(a, bag0) + count(a, bag1);%union-ass: ⊦ bag0 ∪ bag1 ∪ bag2 = (bag0 ∪ bag1) ∪ bag2;%union-com: ⊦ bag0 ∪ bag1 = bag1 ∪ bag0;%count-subset: ⊦ bag0 ⊆ bag1 ↔ (∀ a. count(a, bag0) ≤ count(a, bag1));%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat